A) \[\frac{20}{19!}-2\]
B) \[\frac{21}{20!}-1\]
C) \[\frac{21}{20!}\]
D) none of these
Correct Answer: B
Solution :
[b] \[{{T}_{r}}={{(-1)}^{r}}\frac{{{r}^{2}}+r+1}{r!}\] \[={{(-1)}^{r}}\left[ \frac{r}{(r-1)!}+\frac{1}{(r-1)!}+\frac{1}{r!} \right]\] \[={{(-1)}^{r}}\left[ \frac{1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{1}{(r-1)!}+\frac{1}{r!} \right]\] \[=\left[ \frac{{{(-1)}^{r}}}{r!}+\frac{{{(-1)}^{r}}}{(r-1)!} \right]+\left[ \frac{{{(-1)}^{r}}}{(r-1)!}+\frac{{{(-1)}^{r}}}{(r-2)!} \right]\] \[=\left[ \frac{{{(-1)}^{r}}}{r!}-\frac{{{(-1)}^{r-1}}}{(r-1)!} \right]-\left[ \frac{{{(-1)}^{r-1}}}{(r-1)!}-\frac{{{(-1)}^{r-2}}}{(r-2)!} \right]\] \[=V(r)-V(e-1)\] \[\therefore \sum\limits_{r=1}^{n}{{{T}_{r}}}=V(n)-V(0)=\left[ \frac{{{(-1)}^{n}}}{n!}-\frac{{{(-1)}^{n-1}}}{(n-1)!} \right]-1\] Therefore, the sum of 20 terms is \[\left[ \frac{1}{20!}-\frac{-1}{19!} \right]-1=\frac{21}{20!}-1\]
You need to login to perform this action.
You will be redirected in
3 sec
